Power domination in product graphs∗
Paul Dorbec†
Michel Mollard†
Sandi Klavžar‡
Simon Špacapan§
Abstract
The power system monitoring problem asks for as few as possible measurement
devices to be put in an electric power system. The problem has a graph theory
model involving power dominating sets in graphs. The power domination number
γP (G) of G is the minimum cardinality of a power dominating set. Dorfling and
Henning [2] determined the power domination number of the Cartesian product
of paths. In this paper the power domination number is determined for all direct
products of paths except for the odd component of the direct product of two odd
paths. For instance, if n is even and C a connected component of Pm × Pn , where
m is odd or m ≥ n, then γP (C) = dn/4e. For the strong product we prove that
γP (Pn £ Pm ) = max{dn/3e, d(n + m − 2)/4e}, unless 3m − n − 6 ≡ 4 (mod 8).
The power domination number is also determined for an arbitrary lexicographic
product.
Key words: domination, power domination, electric power monitoring, graph products, paths
2000 Mathematics Subject Classification: 05C69
1
Introduction
Electric power systems need to be continually monitored. One way to fulfill this task
is to place phase measurement units at (carefully) selected locations in the system.
The power system monitoring problem, as introduced in [1], asks for as few as possible
measurement devices to be put in an electric power system.
∗
Supported in part by the Proteus project BI-FR/04-002, by the Ministry of Science of Slovenia
under the grant P1-0297, by the ENS Lyon, and by the CNRS.
†
ERTé Maths à Modeler, GéoD research group, Leibniz laboratory, 46 av. Félix Viallet, 38031
Grenoble CEDEX, France; e-mail: [email protected] (UJF); [email protected] (CNRS)
‡
Department of Mathematics and Computer Science, PeF, University of Maribor, Koroška cesta
160, 2000 Maribor, Slovenia, e-mail: [email protected]. Also with the Institute of Mathematics,
Physics and Mechanics, Jadranska 19, 1000 Ljubljana.
§
University of Maribor, FME, Smetanova 17, 2000 Maribor, Slovenia, e-mail: [email protected]. Also with the Institute of Mathematics, Physics and Mechanics, Jadranska 19, 1000 Ljubljana.
1
The power system monitoring problem has been formulated as a graph-theory domination problem by Haynes, Hedetniemi, Hedetniemi, and Henning in [5]. This problem
is of somehow different flavor than standard domination type problems, since putting a
phase measurement unit into a vertex of a graph can have global effects. For instance,
if an electric power system can be modeled by a path, then a single measurement unit
suffices to monitor the system no matter how long is the path.
Let G be a connected graph and S a subset of its vertices. Then we denote the set
monitored by S with M (S) and define it algorithmically as follows:
1. (domination)
M (S) ← S ∪ N (S)
2. (propagation)
As long as there exists v ∈ M (S) such that
N (v) ∩ (V (G) − M (S)) = {w}
set M (S) ← M (S) ∪ {w}.
The set M (S) is thus obtained from S as follows. First put into M (S) the vertices from
the closed neighborhood of S. Then repeatedly add to M (S) vertices w that have a
neighbor v in M (S) such that all the other neighbors of v are already in M (S). After
no such vertex w exists, the set monitored by S has been constructed.
The set S is called a power dominating set of G if M (S) = V (G) and the power
domination number γP (G) is the minimum cardinality of a power dominating set.
In [5] these concepts have been introduced in a slightly more complicated way by
treating both vertices and edges of a given graph. However, it is easily shown that both
approaches are equivalent in the sense that the power dominating sets correspond. In
fact, this is also (implicitly) observed in [2].
The power domination number has received considerable attention from the algorithmic point of view. As it turned out, the problem is quite difficult—it is NP-complete
even when restricted to bipartite graphs and chordal graphs [5], to planar graphs and
circle graphs [3] as well as to split graphs [3, 9]. On the other hand, the problem has
efficient solutions on trees [5] and on interval graphs [9].
Dorfling and Henning [2] obtained closed formulas for the power domination numbers of grid graphs. This result is in a striking contrast to the fact a determination
of such formulas for the usual domination number of grid graphs is a notorious open
problem, see [6, Section 2.6.2]. Now, a natural description of a grid is as the Cartesian
product of two paths. Besides the Cartesian one, there are three more standard graph
products: the strong, the direct, and the lexicographic product [7]. Hence it is natural
to ask whether the power domination number can also be determined for these products
of paths. In this paper we indeed show to be the case.
We proceed as follows. In the rest of this section we give definitions used in this
paper. In Section 2 we determine the power domination number for direct products
of paths with the exception of the odd components of products of odd paths. For this
2
case an upper bound is given for which we believe to be optimal. Then, in Section 3,
we determine the power domination number for the strong product of paths, except
that in one of the eight cases the power domination number is only bounded. But
also in this case the bounds are almost tight. While it takes quite an effort to obtain
the results for the direct and the strong product of paths, it is rather straightforward
to deal with the lexicographic product. In fact, in the last section we show that the
power domination number can be determined exactly for the lexicographic product of
arbitrary graphs in terms of the (total) domination number of the factors.
For a vertex v of a graph G, let N (v) denote the open neighborhood of v, and for a
subset S of V (G) let N (S) = ∪v∈S N (v) − S. In the same way, the close neighborhood
N [S] of a subset S is the set N [S] = N (S) ∪ S.
All of the four standard graph products constructed from graphs G and H have
vertex set V (G) × V (H). Let (g, h) and (g 0 , h0 ) be two vertices in V (G) × V (H). They
are adjacent in the Cartesian product G ¤ H if they are equal in one coordinate and
adjacent in the other. They are adjacent in the direct product G×H if they are adjacent
in both coordinates. The edge set of the strong product G £ H is E(G ¤ H) ∪ E(G × H).
Finally, (g, h) and (g 0 , h0 ) are adjacent in the lexicographic product G ◦ H if gg 0 ∈ E(G),
or if g = g 0 and hh0 ∈ E(H). All these products are associative, all but the lexicographic
product are also commutative; see [7] for more information on all the products.
Let G ∗ H be any of the four standard graph products. Then the subgraph of G ∗ H
induced by {g} × V (H) is called a H-fiber and denoted g H. Similarly one defines the
G-fiber Gh for a vertex h of H. If ∗ is the Cartesian, the strong, or the lexicographic
product, then fibers are isomorphic to the corresponding factors. For the direct product
the fibers are discrete, that is, they induce edgeless graphs.
Finally, for the path Pm we will always assume that V (Pm ) = {0, . . . , m − 1}, where
vertices are adjacent in the natural way. Moreover, the vertices of Pm ∗ Pn will be
denoted by (i, j) where 0 ≤ i < m and 0 ≤ j < n .
2
The direct product
The direct product Pm × Pn is not connected. If m or n is even Pm × Pn consists of
two isomorphic connected components, otherwise it has two non-isomorphic connected
components, see [10], cf. also [4, 8]. We call the component of Pm × Pn containing
(0, 0) the even component and the other component the odd component.
In this section we determine the power domination number of the direct product
of two paths, except for the odd component of the product of two odd paths. In the
first subsection we treat the case when at least one of the paths is even. In this case
γP (C) = dn/4e, where C is a connected component of Pm × Pn with n even and m odd
or m ≥ n. In the subsequent subsection we follow with products of odd paths proving
that for the even component C, γP (C) = d(m + n)/6e. We close by construction a
power dominating set of the odd component of such products. We conjecture that the
3
construction is optimal.
Let p1 : Pm × Pn → Pm and p2 : Pm × Pn → Pn be the natural projections onto the
first and the second factor of Pm × Pn , respectively. Then:
Lemma 2.1 Let S be a power dominating set of a connected component C of Pm × Pn .
If m is odd, then for every subpath P ⊆ Pn of length 3, p2 (S) ∩ P 6= ∅.
Proof. Suppose on the contrary that there exists a subpath P = x1 x2 x3 x4 of length
3, such that p2 (S) ∩ P = ∅.
x3 ∩ C cannot be power dominated since
Assume C is even. Then if x1 is even, Pm
x3 ∩ C has two neighbors in P x3 . Similarly we infer
any neighbor of a vertex from Pm
m
x
2
that if x1 is odd then Pm ∩ C cannot be power dominated, a contradiction. Arguments
¤
for C being odd are analogous.
2.1
At least one factor is even
To solve the case when at least one factor is an even path, we begin with two lemmas.
Lemma 2.2 Let m and n be even and let S be a power dominating set of a connected
component C of Pm ×Pn . If there is a subpath P ⊆ Pm of length 3, such that P ∩ p1 (S) =
∅ then for every subpath Q ⊆ Pn of length 3, Q ∩ p2 (S) 6= ∅.
Proof. Suppose on the contrary that P ⊆ Pm and Q ⊆ Pn are paths of length 3 with
p1 (S) ∩ P = p2 (S) ∩ Q = ∅. Assume that C is even and that all vertices from
F = {(i, j) | i ∈
/ P, j ∈
/ Q} ∩ C
are monitored. We claim that all monitored vertices are contained in the neighborhood
of F . Suppose that (u, v) is a vertex from F , such that all but one neighbors of (u, v)
are contained in F (and thus monitored). Then clearly (u, v) lies on the boundary of
Pm × Pn . Hence all the newly obtained monitored vertices from C \ F do not lie on the
boundary of Pm × Pn , and therefore have at least two neighbors not monitored. Thus
all monitored vertices are contained in the neighborhood of F . Since N [F ] 6= C we find
¤
that S is not a power dominating set of C.
Lemma 2.3 Let m be even and let C be a connected component of Pm × Pn . If all the
u ∪ P v ) ∩ C are monitored, where u and v are adjacent vertices of P ,
vertices from (Pm
n
m
then C is power dominated.
u ∩ C and P v ∩ C contain a vertex that has exactly
Proof. Since m is even, both Pm
m
u ∪ P v ) ∩ C. It follows easily that the two neighboring fibers
one neighbor not in (Pm
m
(or the neighboring fiber if there is only one such) are monitored. Induction completes
the argument.
¤
4
Theorem 2.4 Let n be even and C be a connected component of Pm × Pn . If m is odd
or m ≥ n, then γP (C) = dn/4e.
Proof. We first consider the case when m is odd. Then, since n is even, either all
1 ∩ C have two neighbors in P 0 ∩ C or all vertices from P n−2 ∩ C
vertices from Pm
m
m
n−1 ∩ C. It follows that any power dominating set S ⊆ C has
have two neighbors in Pm
0 , P 1 , P n−2 and P n−1 .
nonempty intersection with one of the fibers Pm
m
m
m
By Lemma 2.1 a power dominating set S of C has the property that for any subpath
1 ∩S
P ⊆ Pn of length 3, P ∩ p2 (S) 6= ∅. Without loss of generality assume that Pm
0 as well as the fibers P n−2 and P n−1 (but not P n−3 )
is nonempty. Since the fiber Pm
m
m
m
might have empty intersection with S and at least every fourth fiber has nonempty
intersection with S we find that
|S| ≥ |{1, 1 + 4, . . . , 1 + 4 d(n − 4)/4e}| = dn/4e .
Hence γP (G) ≥ dn/4e. To prove the reverse inequality we need to construct a power
dominating set S of C with |S| = dn/4e. We may without loss of generality assume
that C is even. Then let:
S = {(1, 1 + 4α) | α = 0, . . . , dn/4e − 1} .
(1)
Observe that all the vertices from 0 Pn are monitored since for every vertex from 0 Pn
there is a neighboring vertex from S which dominates it; see the left-hand side of Fig. 1.
Next observe that also all the vertices from 1 Pn are monitored, since for every vertex
from 1 Pn there is a vertex from 0 Pn with all but one vertices in S. Since the consecutive
fibers 0 Pn and 1 Pn are monitored and n is even, Lemma 2.3 implies that S is a power
dominating set of C. This completes the proof for m being odd.
Let now m be even, m ≥ n. Let S be a power dominating set of Pm × Pn . By
Lemma 2.2, there is either no subpath of Pm of length 3 disjoint with p1 (S) or there
is no subpath of Pn of length 3 disjoint with p2 (S). Hence |S| ≥ bn/4c. This is the
desired lower bound if n = 4k.
If n = 4k + 2 and |S| = k then either there is a subpath P ⊆ Pn of length 3 disjoint
0 , P 1 and P 2 , or
with p2 (S), or the first three or the last three Pm -fibers (that is, Pm
m
m
n−3
n−2
n−1
Pm , Pm and Pm ) have empty intersection with p2 (S). In the first case there is no
subpath of Pm of length 3 disjoint with p1 (S), therefore in this case m = n. But this
means that S is disjoint with fibers 0 Pn ,1 Pn and 2 Pn (or m−3 Pn ,m−2 Pn and m−1 Pn ).
Therefore 0 Pn is not monitored (see the right-hand side of Fig. 1), a contradiction. The
0 , P 1 and P 2 have empty intersection with p (S). If also 0 P ,1 P
second case is when Pm
2
n
n
m
m
2
0 are not monitored.
and Pn have empty intersection with p1 (S), then 0 Pn and Pm
Otherwise there is a subpath of Pm of length 3 disjoint with p1 (S). This case leads to
a contradiction similarly as shown on Fig.1. Thus if n = 4k + 2 and |S| = k, then S
cannot be a power dominating set. Hence in either of the cases |S| ≥ dn/4e. Finally,
an optimal power dominating set S of cardinality n/4 for the case when m is even is
again defined by (1).
¤
5
S
Pn
Pn
S
Pm
Pm
Figure 1: An optimal power dominating set of Pm × Pn is shown on the left. The
maximal set of monitored vertices is marked with broken line on the right.
Note that Theorem 2.4 holds also for n = 2. Indeed, a connected component of
Pm × P2 is isomorphic to Pm whence its power domination number is 1.
2.2
Both factors are odd
We continue with the odd times odd case. For this case we introduce the following
concept on a grid Pm ¤ Pn .
Consider an initial set S of vertices. Let D be initialized by D = S. As long as
there is a vertex v in V \ D such that v has at least two neighbors in D, add v to D.
If at the end of the process D = V (Pm ¤ Pn ), then we call the set S a life winning set
of Pm ¤ Pn , LWS, for short. Let us call this concept the life-like game.
Theorem 2.5 For any positive integers m, n, there exists a LWS set of cardinality
d m+n
2 e and it is minimum.
Proof. Suppose m and n are odd. Then let S consists of vertices (0, 0), (2, 0), . . . , (m −
1, 0), (m − 1, 2), (m − 1, 4), . . . , (m − 1, n − 1). In the other cases S is constructed
similarly. It is straightforward to verify that S is a LWS set of the desired cardinality.
Let us prove that SPis minimum. For X ⊆ V (Pm ¤ Pn ) let Π(X) be the perimeter
of X, that is, Π(X) = v∈X (4 − |N (v) ∩ X|). If D0 is a subset of V (Pm ¤ Pn ) obtained
by propagation from D, then Π(D0 ) ≤ Π(D). Indeed, any vertex added to D decreases
6
Π(D) by at least 2 and adds at most 2 to Π(D). At the end of the propagation,
Π(D) = 2(m + n). At the beginning, Π(S) ≤ 4|S|. Hence, |S| ≥ (m + n)/2.
¤
The above theorem might be of independent interest, but for our purposes it is
important that during the propagation process the perimeter of the monitored set does
not increase.
In the following we will consider the direct product Pm × Pn and at the same time
the Cartesian product Pm ¤ Pn , where the vertex sets of both products are the same.
Lemma 2.6 Let m, n be odd integers and P a power dominating set of the even component of Pm × Pn . Let S be the set monitored by P after the domination step of the
algorithm, then S is a LWS of Pm ¤ Pn .
Proof. We first remark that all the vertices of the even component either have both
their coordinates odd or both even. We will call them odd and even vertices, respectively. Since both m and n are odd, the set of even vertices is a LWS of Pm ¤ Pn .
Let M and D, initialized to S, be the sets that are propagating for power domination
and life-like game respectively. We will prove by induction that any even vertex in M
is in D. From the choice M = D = S, this condition is initially true.
2i
2i+1 2i+2
2i
w
2i+1 2i+2
w
2j
2j
v
v
2j+1
2j+1
2j+2
2j+2
Figure 2: Propagation from an odd vertex to an even
Suppose an even vertex w is added to M . Then, it is monitored by propagation
from an odd vertex v. Any odd vertex has 4 neighbors, all even. So all the 3 other
neighbors of v are monitored and belong to D. In Fig. 2, we show how w is thus also
put in D by the rule of the life-like game.
¤
Theorem 2.7 Let m and n be odd and C the even component of Pm × Pn . Then
¼¾
½l m »
m+n
n
,
.
γP (C) = max
4
6
0 or P 1 , and a
Proof. Since any power dominating set of C contains a vertex from Pm
m
n−1
n−2
vertex from Pm or Pm , Lemma 2.1 implies that γP (C) ≥ n/4.
7
Let P be a power dominating set of Pm × Pn . Let S be the set monitored after
the domination step. Let us consider the life-like game starting on S. Every vertex
from P contributes at most 5 vertices to S which propagate to a 3 × 3 box in the grid.
Denoting the obtained set S 0 we infer that every vertex in P contributes at most 12 to
Π(S 0 ). The set S 0 is a LWS of the grid, so 12 |P | ≥ 2(m + n). (Here we have used the
fact that during the propagation process the perimeter of the monitored set does not
increase.) Therefore |P | ≥ d(n + m)/6e.
To establish the upper bound we construct a corresponding power dominating set
P of C as follows. Suppose first that m ≤ n ≤ 2m and set
S1 = {(1 + 2i, 1 + 4i) | 0 ≤ i ≤ (2n − m − 3)/6}
S2 = {((2n − m)/3 + 4j, (4n − 2m − 3)/3 + 2j) | 1 ≤ j ≤ (2m − n − 3)/6} .
Suppose that n + m ≡ 0 (mod 6) (the perfect case). Then set P = S1 ∪ S2 . Otherwise,
construct P from a maximal perfect case contained in Pm × Pn by adding the vertex
(m − 2, n − 2) to it. It is straightforward to verify that in all cases P is a power
dominating set. Moreover, ((2n − m − 3)/6 + 1) + (2m − n − 3)/6 = (m + n)/6, hence
|P | = d(m + n)/6e.
Let n > 2m. Then set
S1 = {(1 + 2i, 1 + 4i) | 0 ≤ i ≤ (m − 3)/2}
S2 = {(1, 1 + 4j) | (m − 3)/2 < j ≤ (n − 3)/4} .
Now the perfect case is when n ≡ 3 (mod 4), in which case we set P = S1 ∪ S2 .
Otherwise, proceed as above by adding (m − 2, n − 2) to the power dominating set of a
maximal perfect case contained in Pm ×Pn . Since (m−3)/2+1+((n−3)/4−(m−3)/2) =
¤
(n + 1)/4, in this case |P | = dn/4e.
The construction for the first case from the previous proof is illustrated in Fig. 3
with black dots. Moreover, another optimal power dominating set is presented with
black squares in order to illustrate that such a set can be unusual.
Theorem 2.8 Let m and n be odd and C the odd component of Pm × Pn . Then
½»
¼ »
¼¾
n−2
m+n−2
γP (C) ≤ max
,
.
4
6
Proof. We give a construction of a power dominating set of the given cardinality.
Suppose m ≤ n ≤ 2m + 2 and m + n − 2 ≡ 0 (mod 6) (the perfect case), then the
construction is as follows. Let
S1 = {(3 + 2i, 4i) | 1 ≤ i ≤ (2n − m − 7)/6} and
S2 = {(2n − m − 1)/3 + 4j, (4n − 2m − 11)/3 + 2j) | 1 ≤ j ≤ (2m − n − 7)/6} .
8
Figure 3: Two optimal power dominating sets of the even component of P17 × P19
Then P = S1 ∪ S2 ∪ {(2, 1), (m − 2, n − 3)} is a power dominating set of C. Since
(2n − m − 7)/6 + (2m − n − 7)/6 + 2 = (m + n − 2)/6, this construction has the claimed
cardinality.
In the non perfect case, construct S1 and S2 for a minimal perfect case containing
Pm × Pn . Add to these sets the vertices (2, 1) and (m − 2, n − 3) (thus, only this last
vertex differs). This set clearly has the cardinality d(m + n − 2)/6e.
If n > 2m + 2 and n ≡ 1 (mod 4) then let
S1 = {(1 + 2i, 2 + 4i) | 0 ≤ i ≤ (m − 3)/2} and
S2 = {(1, 2 + 4j) | (m − 3)/2 ≤ j ≤ (n − 5)/4} .
In the non perfect case (n ≡ 3 (mod 4)), construct S1 and S2 for Pm × Pn+2 and it will
form a P DS of Pm × Pn . Since (m − 3)/2 + 1 + ((n − 5)/4 − (m − 3)/2) = (n − 1)/4,
in this case |P | = d(n − 2)/4e.
¤
We think that the upper bound of Theorem 2.8 is optimal. However, proving that
this is the case seems harder than it was for the even component.
9
3
The strong product
In this section we prove the following theorem.
Theorem 3.1 Let n ≥ m ≥ 1. Then
γP (Pn £ Pm ) = max
nl n m l n + m − 2 mo
,
3
4
unless 3m − n − 6 ≡ 4 (mod 8) in which case
max
nl n m l n + m − 2 mo
nl n m l n + m − 2 m
o
,
≤ γP (Pm £ Pn ) ≤ max
,
+1 .
3
4
3
4
In the first subsection we prove the lower bounds, while in the subsequent two subsections corresponding power dominating sets are constructed. It is easily verified that
the result holds if n < 4 and m < 4. Hence in the rest of this section we assume that
n ≥ m ≥ 4.
3.1
Proof of the lower bound
Lemma 3.2 γP (Pm £ Pn ) ≥
lnm
.
3
Proof. Let S be a power dominating set of Pm £ Pn . Suppose there is a Pm -fiber that
does not contain any vertex of N [S]. Then, the vertices of this Pm -fiber have to be
monitored during the propagation step. Consider the first vertex v of this Pm -fiber
that is monitored, and let w be the vertex from whom it is monitored. w has at least
one other neighbor in the same Pm -fiber than v. This neighbor is not yet monitored,
so the propagation cannot happen. Thus S must dominate at least one vertex of each
Pm -fiber which implies that γP (Pm £ Pn ) ≥ dn/3e.
¤
Lemma 3.3 γP (Pm £ Pn ) ≥
n+m−2
.
4
Proof. Let S be a minimum power dominating set of Pm £ Pn . We can suppose that
S contains no vertices on the boundary (that is, the vertices of Pm £ Pn of degree 3
or 5), since any closed neighborhood of a vertex on the boundary is contained in the
closed neighborhood of a vertex not on the boundary.
Let M be any set of monitored vertices of Pm £ Pn during a propagation step of
the algorithm. Let B(M ) be the set of vertices of M that have less than 8 neighbors
in M . We claim that during the propagation step, |B(M )| cannot increase.
Suppose that during the propagation step we monitor a new vertex w, where w is
the only neighbor not in M of some vertex v from M . Then w may now be added to
B(M ). We distinguish two cases.
10
Case 1. the degree of v is 8.
Then v now has all its 8 neighbors in M , so v is removed from B(M ). Some other
vertices may be removed from B(M ), if w was their eighth neighbor to enter B(M ),
but none other may be added. So |B(M )| can not be increasing.
Case 2. the degree of v is less than 8.
In this case we check that w could have been monitored with a propagation from a
vertex not on the boundary. The cases to be considered are shown in Fig. 4.
(a)
!"! "! "!
!!" !" !"
(c)
#$# $# $#
#$# $# $#
;;;;
;<; <; <;
<<<
=>= >= >=
=>= >= >=
(e)
'(' (' ('
'(' (' ('
)*) *) *)
)*) *) *)
++++
+,+ ,+ ,+
,,,
%&% &% &%
%&% &% &%
-.- .- .-.- .- .-
787 87 87
787 87 87
(b)
?@? @? @?
??@ ?@ ?@
DC DC DC
CD CD CD
EFE FE FE
EFE FE FE
BA BA BA
BABA BA
(d)
9:9 :9 :9
99: 9: 9:
(f)
343 43 43
334 34 34
565 65 65
565 65 65
121 21 21
112 12 12
/0/ 0/ 0/
/0/ 0/ 0/
Figure 4: Propagation from the border
In case (a), we can consider that the propagation comes from the vertex in a box, which
has 8 neighbors. Cases (b), (e), and (f) cannot occur, because the vertices in circles
must have been monitored with some propagation, which is impossible. In cases (c)
and (d), the two vertices in circles must have been monitored by propagation, but none
could have been before the other was. The case of the corner is induced by cases e and
f . This proves the claim.
Hence |B(M )| cannot increase during a propagation step of the algorithm. After
the domination step of the algorithm, |B(M )| is not greater than 8 |S| . Moreover,
at the end of the algorithm, when all the vertices are monitored, |B(M )| is equal to
2n + 2m − 4. So we must have |S| ≥ (n + m − 2)/4.
¤
3.2
A construction for n ≥ 3m − 6
In this subsection we construct power dominating sets for the case when one path is
much longer than the other, more precisely, when n ≥ 3m − 6. Then
lnm
3
≥
l 3n
12
+
3m − 6 m l n + m − 2 m
=
,
12
4
11
hence the power domination number depends only of the size of the longer path. Thus
we need to construct a power dominating set S of cardinality dn/3e. We do this as
follows. Set S = S1 ∪ S2 , where
S1 = {(1 + i, 1 + 3i) | 0 ≤ i ≤ m − 3}
and
½
S2 =
{(1, 3m − 9 + 3j) | 1 ≤ j ≤ d(n − 3m + 6)/3e}; n ≡ 1 (mod 3),
{(1, 3m − 8 + 3j) | 1 ≤ j ≤ d(n − 3m + 6)/3e}; otherwise.
The construction is illustrated in Fig. 5.
S2
S1
n mod 3 = 0
n mod 3 = 1
Figure 5: Optimal power dominating sets when one path is relatively long
It is easy to check that the set S will power dominate the whole strong product.
Moreover, |S| = |S1 | + |S2 | = m − 2 + d(n − 3m + 6)/3e = dn/3e.
3.3
A construction for n ≤ 3m − 6
In this case m and n are comparable. Since dn/3e ≤ d(n + m − 2)/4e we need to
construct power dominating sets S of cardinality d(n + m − 2)/4e. We start with the
case when 8 divides 3m − n − 6, and modify it in all the other cases.
12
Case 1. 3m − n − 6 ≡ 0 (mod 8).
Set S = S1 ∪ S2 , where
S1 = {(1 + 3i, 1 + i) | 0 ≤ i ≤ (3m − n − 6)/8}
S2 = {(1 + 3(3m − n − 6)/8 + j, 1 + (3m − n − 6)/8 + 3j) |
1 ≤ j ≤ (n + m − 6)/4 − (3m − n − 6)/8} .
The construction is illustrated in Fig. 6.
S2
S1
Figure 6: Power dominating sets of products of path of similar length
For coherence of the above construction we should remark that:
•
3m−n−6
8
•
n+m−6
4
and
−
n+m−6
4
3m−n−6
8
=
=m−3−
3n−m−6
8
3m−n−6
4
are integers.
is positive since n ≥ m and n ≥ 4.
• The coordinates of the last vertex which is furthest down and right are (m−2, n−
2), so all the vertices are in the graph.
13
• The set S does not contain any vertex on the border of the grid. (This remark
will be very useful in the following.)
The set S contains (n + m − 2)/4 vertices and is therefore optimal.
Case 2. 3m − n − 6 ≡ 1 (mod 8).
In this case 3m − (n + 1) − 6 ≡ 0 (mod 8). Then we construct the power dominating
0 from P £ P
set S of Pm £ Pn+1 from Case 1. Removing the fiber Pm
m
n+1 we get a
graph G = Pm £ Pn . Since S does not contain any border vertex, the set S restricting
to G does not change. Consequently, |S| = (n + 1 + m − 2)/4 = d(n + m − 2)/4e.
Case 3. 3m − n − 6 ≡ 2 (mod 8).
Then 3m − (n + 2) − 6 ≡ 0 (mod 8), hence construct the power dominating set S of
0 and
Pm £ Pn+2 as in Case 1 and restrict it to the graph Pm £ Pn+2 with the fibers Pm
n+1
Pm removed. Now |S| = (n + 2 + m − 2)/4 = d(n + m − 2)/4e.
Case 4. 3m − n − 6 ≡ 3 (mod 8).
Since 3(m + 2) − (n + 1) − 6 ≡ 0 (mod 8) we take the power dominating set S of
0 . Then |S| = (n + 1 + m + 2 −
Pm+2 £ Pn+1 and remove the fibers 0 Pn , m+1 Pn , and Pm
2)/4 = d(n + m − 2)/4e.
Case 5. 3m − n − 6 ≡ 4 (mod 8).
As 3(m + 2) − (n + 2) − 6 ≡ 0 (mod 8) we proceed as before, removing the fibers 0 Pn ,
m+1 P , P 0 , and P n+1 . Then |S| = (n + 2 + m + 2 − 2)/4 = d(n + m − 2)/4e + 1. Note
n
m
m
that in this special case we miss the lower bound by 1.
Case 6. 3m − n − 6 ≡ 5 (mod 8).
Then 3(m + 1) − n − 6 ≡ 0 (mod 8) so we construct the power dominating set S of
Pm+1 £ Pn and remove the fiber 0 Pn . Then |S| = (n + m + 1 − 2)/4 = d(n + m − 2)/4e.
Case 7. 3m − n − 6 ≡ 6 (mod 8).
Then 3(m + 1) − (n + 1) − 6 ≡ 0 (mod 8) and we start from the power dominating set S
0 . Now |S| = (n + 1 + m + 1 − 2)/4 =
of Pm+1 £ Pn+1 , and remove the fibers 0 Pn and Pm
d(n + m − 2)/4e.
Case 8. 3m − n − 6 ≡ 7 (mod 8).
Then 3(m + 1) − (n + 2) − 6 ≡ 0 (mod 8). Take the power dominating set S in
0 , and P n+1 . We have |S| = (n + 2 + m +
Pm+1 £ Pn+2 and remove the fibers 0 Pn , Pm
m
1 − 2)/4 = d(n + m − 2)/4e.
4
The lexicographic product
The remaining standard graph product to consider is the lexicographic one. For it
the power domination problem is much easier. In fact we will determine the power
domination number for any lexicographic product in terms of the domination number
and the total domination number of its factors.
14
Recall that for a graph G = (V, E), a set S ⊆ V is a dominating set if each vertex in
V \S is adjacent to at least one vertex of S. If in addition each vertex of S has a neighbor
in S, then S is called a total dominating set. The domination (resp. total domination)
number γ(G) (resp. γt (G)) of G is the minimum cardinality of a dominating (resp.
total dominating) set. Then we have:
Theorem 4.1 For any nontrivial graphs G and H,
½
γ(G); γP (H) = 1 ,
γP (G ◦ H) =
γt (G); γP (H) > 1 .
Proof. Suppose γP (H) = 1. Let {v} be a power dominating set of H and D a
dominating set of G. Then {v} × D is a power dominating set of G ◦ H. Indeed, for
any vertex u of G, if u ∈
/ D, any vertex of u H is in the neighborhood of (u, v), so it
is monitored. If u ∈ D, any neighbor of a vertex of u H not in u H is monitored, and
{(u, v)} ∪ N (u, v) is monitored. So since {v} is a power dominating set of H, the fiber
is monitored. Therefore γP (G ◦ H) ≤ γ(G).
Assume that there is a power dominating set S of G ◦ H that contains less than
γ(G) vertices. Then there is an H-fiber u H that contains no vertex of S ∪ N (S). So
the vertices of u H are monitored by the propagation. The first vertex of u H that is
monitored must be the only neighbor not monitored by some vertex not in u H. But
this vertex is also adjacent to all the other vertices of u H, which are not monitored
yet. Since H is not trivial this implies that there can be no propagation on u H. Hence
γP (G ◦ H) ≥ γ(G).
Suppose now γP (H) > 1. Let D be a total dominating set of G. Then for any
vertex v of H, D × {v} is a dominating set of G ◦ H and so a power dominating set of
H of cardinality |D| = γt (G). Thus γP (G ◦ H) ≤ γt (G).
Let P be a minimum power dominating set of G ◦ H. Suppose that there exists an
H-fiber u H such that for any neighbor v of u in G, v H ∩ P is empty. Thus, the vertices
monitored by P in u H are exactly the vertices monitored by P ∩ u H. Since there is no
power dominating set of H of cardinality 1, P ∩ u H contains at least two vertices. Let
P 0 be obtained from P by removing all vertices from P ∩u H but one and adding an
arbitrary vertex of v H, where v is a neighbor of u in G. Then P 0 is a power dominating
set of G ◦ H with |P 0 | ≤ |P |. By repeating this process we construct a minimum power
dominating set P ∗ of G ◦ H such that any vertex u ∈ V (G) has a neighbor v in G whose
H-fiber v H contains a vertex of P ∗ . So the set {v ∈ V (G) | v H ∩ P ∗ 6= ∅} is a total
dominating set of G and we conclude that γP (G ◦ H) = |P ∗ | ≥ γt (G).
¤
Acknowledgement
We wish to thank Sylvain Gravier for introducing us to the concept of the power
domination and for useful conversations.
15
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